Differential Equations Solutions 43

Differential Equations Solutions 43 - Chapter 10 Solutions:...

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Chapter 10 Solutions: Numerical Methods for Constrained Optimization CHALLENGE 10.1. The graphical solutions are left to the reader. (a) The optimality condition is that the gradient should be zero. We calculate g ( x )= · 2 x 1 x 2 +5 8 x 2 x 1 +3 ¸ , H ( x · 2 1 18 ¸ . Since H is positive deFnite (Gerschgorin theorem), f ( x ) has a unique minimizer satisfying g ( x 0 ,so H ( x ) x = · 5 3 ¸ , and therefore x =[ 2 . 8667 , 0 . 7333] T . (b) Using a Lagrange multiplier we obtain L ( x x 2 1 +4 x 2 2 x 1 x 2 x 1 x 2 +6 λ ( x 1 + x 2 2) .
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.

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